Question

Let $k$ be the number of sets of $2$ and $3$ (different) numbers that can be formed by using numbers between $0$ and $180$ (both including) so that $60$ is their average.Find sum of largest and smallest digits of $k$ ?

Answer 1

$\left(i\right)$ Set of $2$ numbers:
Let $a$ and $B$ be $2$ numbers
$\frac{a+b}{2}=60\Rightarrow a+b=120$
$a$ and $b$ both can't be equal to or greater than $60$ $(\because 60$ can't be used twice $)$
Let $0\le a\le 59$ and $61\le b\le 120$
The total no. of ways in which $a$ can be chosen $={}^{60}{C}_1=60$
The value of $b$ depends on the value of $a$ and there is $1$ value of $b$ corresponding to $1$ of $a$.
$\therefore$ Total no. of sets having $2$ numbers $=60$
$\left(ii\right)$ Set of $3$ numbers:
Let $a,b,c$ be the three numbers
Then $\frac{a+b+c}{3}=60\Rightarrow a+b+c=180$.
Case I: Let $0\le a\le 59,0\le b\le 59$ and $c\ge 60$
$a$ can be chosen in ${}^{60}{C}_1=60$ ways.
$b$ can be chosen in ${}^{59}{C}_1=59$ ways. $(\because b$ can't use the value of $a)$
$\therefore$ no. of ways in which $a$ and $b$ can be chosen $=60\times 59=3540$
Now $1\le a+b\le 117$ and there is only one value of $c$ for $1$ value of $a+b$ so that $a+b+c=180$
$\therefore$ No. of ways in which $a,b,c$ can be chosen $=60\times 59=3540$.
Case II: $a=60$, $\therefore b+c=120$
The no. of ways in which $b$ and $c$ can assume values $=60$ {from (i)}
$\therefore$ no. of ways in which $a,b,c$ can be chosen $=60$
Case III: $61\le a\le 90,61\le b\le 90$ and $c<60$
$a$ can assume values in ${}^{30}{C}_1=30$ ways
$b$ can assume values in ${}^{29}{C}_1=29$ ways
The value of $c$ depends on the value of $a$ and $b$
$\therefore$ no. of ways in which $a,b,c$ can be chosen $=30\times 29=870$
$\therefore$ no. of ways in which sets of $3$ numbers can be chosen
$=3540+60+870=4470$
$\therefore$ Total no. of ways in which sets of $2$ and $3$ numbers can be chosen
$=4470+60=4530$.